Because you need to join the dirname
with x
, os.listdir()
just lists the contents directly, the contents do not have full path.
Example -
[x for x in os.listdir('dirname') if os.path.isfile(os.path.join('dirname',x))]
When the full path is not given, os.path.isfile()
searches in the current directory, hence when you give '.'
to os.listdir()
you get a correct list back.
Example -
Lets say some folder - /a/b/c
- has files - x
and y
in it.
when you do - os.listdir('/a/b/c')
, the list returned looks like -
['x','y']
Even if you give absolute path inside os.listdir()
, the files returned in the list would have relative path to the dir. You would manually need to join dir and x
to get the correct results.
In your third example, it does not work because os.path.abspath()
also works with current directory, so if you do something like -
os.path.abspath('somefile')
The result produced would be - /path/to/current/directory/somefile
- it does not validate if that is a real file/dir or not.
It is clearly stated in the documentation (Emphasis mine) -
os.path.abspath(path)
Return a normalized absolutized version of the pathname path. On most platforms, this is equivalent to calling the function normpath() as follows: normpath(join(os.getcwd(), path))
.
where os.getcwd()
returns the path to current working directory.
dirname
folder? – Lussierdirname
just an example – Mercie